设空间曲线为：x−b1a1=y−d1c1=z1 \frac{x-b_1}{a_1} = \frac{y-d_1}{c_1} = \frac{z}{1}a1​x−b1​​=c1​y−d1​​=1z​x−b2a2=y−d2c2=z1 \frac{x-b_2}{a_2} = \frac{y-d_2}{c_2} = \frac{z}{1}a2​x−b2​​=c2​y−d2​​=1z​当两条直线存在交点时，交点(x0,y0,z0)(x_0, y_0, z_0)(x0​,y0​,z0​)同时满足两条曲线方程，此时：x0=a1z0+b1=a2z0+b2 x_0 = a_1 z_0 + b_1 = a_2 z_0 +b_2x0​=a1​z0​+b1​=a2​z0​+b2​y0=c1z0+d1=c2z0+d2 y_0 = c_1 z_0 + d_1 = c_2 z_0 +d_2y0​=c1​z0​+d1​=c2​z0​+d2​联立两式可得：z0=b2−b1+d1−d2a1−a2+c2−c1 z_0 = \frac{b_2 - b_1 + d_1 - d_2}{a_1 - a_2 + c_2 - c_1}z0​=